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If the system of linear equations \left\{ \begin{align} & 3x-9y=-6 \\ & \dfrac{1}{2}x-\dfrac{3}{2}y=c \\ \end{align} \right. has infinitely many solutions, then the value of contact c is(a) -6(b) -3(c) -2 (d) -1

Last updated date: 14th Sep 2024
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Hint: To begin with, we will first understand what a system of linear equations in two variables means in a graphical 2D plane and what are the conditions to get a unique solution, to get no solutions and to get infinitely many solutions. Then we will apply the condition for infinitely many solutions on the given set of linear equations and try to find the value of c.

In the 2D graphical space, a linear equation in two variables represents a straight line.
If two linear equations are given to use, this means the 2D graphical space has two random lines simultaneously.
There can be 3 ways in which these two lines can occupy place in the 2D space.
Intersecting: If two straight lines are intersecting, they will intersect only at one point. Thus, on solving equations of two intersecting lines, we get a single unique solution, a point which will be common for both the lines.
Parallel: if two straight lines are parallel, they will not intersect at any point. Thus, on solving the two equations, we will not get any solution.
Coincident: If two straight lines are coincident, every point on the line will be common for both lines. Thus, there will be infinitely many solutions.
Let ax + by = c and px + qy = r are any two lines.
For coincident (infinitely many solutions): $\dfrac{a}{p}=\dfrac{b}{q}=\dfrac{c}{r}$
Now, we are given that system of equations \left\{ \begin{align} & 3x-9y=-6 \\ & \dfrac{1}{2}x-\dfrac{3}{2}y=c \\ \end{align} \right. has infinitely many solutions.
Thus, using the condition for coincident lines:
\begin{align} & \Rightarrow \dfrac{3}{\dfrac{1}{2}}=\dfrac{-9}{-\dfrac{3}{2}}=\dfrac{-6}{c} \\ & \Rightarrow \dfrac{6}{1}=\dfrac{-6}{c} \\ & \Rightarrow c=-1 \\ \end{align}
Therefore, the value of c = -1
So, the correct answer is “Option D”.

Note: If ax + by = c and px + qy = r are two linear equations, then condition for intersecting (unique solution) is $\dfrac{a}{p}\ne \dfrac{b}{q}\ne \dfrac{c}{r}$ and for parallel (no solution) is $\dfrac{a}{p}=\dfrac{b}{q}\ne \dfrac{c}{r}$.